Strict Ordering on Integers is Well-Defined
Theorem
Let $\eqclass {a, b} {}$ denote an integer, as defined by the formal definition of integers.
Let:
| \(\ds \eqclass {a, b} {}\) | \(=\) | \(\ds \eqclass {a', b'} {}\) | ||||||||||||
| \(\ds \eqclass {c, d} {}\) | \(=\) | \(\ds \eqclass {c', d'} {}\) |
Then:
| \(\ds \eqclass {a, b} {}\) | \(<\) | \(\ds \eqclass {c, d} {}\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \eqclass {a', b'} {}\) | \(<\) | \(\ds \eqclass {c', d'} {}\) |
Proof
This is a direct application of the Extension Theorem for Total Orderings.
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 5$: The system of integers